Trigonometry and quadratics : Possible mismatch?

Question

There’s this problem I came across, gives me an invalid answer by using general quadratic formula. Wonder why?

$2\sin^2{x} -5\cos{x} -4 =0 $

Here’s what I did:

$2\sin^2{x} -5\cos{x} -4 =0 $

$2(1-\cos^2{x}) - 5 \cos{x} - 4 = 0$

$2 \cos^2{x} + 5 \cos{x} + 2 =0$

This is a quadratic function of $\cos x$, thus,

$\cos{x} = (-5 + 3)/4$ or $\cos{x}= (-5 - 3)/4$

But, the answer given is $\cos{x}=-\frac{1}{2} $ and WolframAlpha says the same but doesn’t show how to do it.

What did I do wrong?

Update: Thank you very much, everyone. Turns out that I wrote the squareroot of 9 as squareroot of 3. My bad

Answer 1

To solve $2y^2+5y+2=0$ either note $$y=\frac {-5\pm \sqrt {25-16}}4$$ so that $y=-2$ or $y=-\frac 12$, and only the second qualifies as a possible value for a cosine, or note the factorisation $$2y^2+5y+2=(2y+1)(y+2)$$

Somehow you have ended up with $\sqrt 3$ rather than $3$ in your computation of the quadratic formula.

Answer 2

Yes from here (you are right) we have:

$$2\sin^2x -5\cos x -4 =0 \implies 2\cos^2 x+5\cos x +2=0$$

that is by quadratic formula (here was your mistake):

$$\cos x = \frac{-5\pm \sqrt{25-16}}{4}= \frac{-5\pm 3}{4}\implies \cos x=-\frac12$$

Answer 3

Simple factorization..

$$2 \cos^2 x + 5 \cos x + 2 =0$$ $$(2 \cos^2 x + 4 \cos x) +(\cos x + 2) =0$$ $$2 \cos x(\cos x +2)+( \cos x + 2) =0$$ $$(\cos x +2)( 2\cos x + 1) =0$$ $$\implies ( 2\cos x + 1) =0$$ $$\implies \cos x =-\frac 12$$